64. [Mcoi2018]终末之诗(上)

无限次幂取模算法

最新推荐文章于 2025-06-06 09:47:56 发布

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原文链接：http://www.cnblogs.com/ZUTTER/p/10229858.html

本文探讨了如何求解无限次幂表达式\(k^{k^{k^{k^{...}

Description

求出\(k^{k^{k^{k^{...}}}} \pmod p\) 的结果

扩展欧拉定理：\[a^x=a^{min(x,x\%\varphi(p)+\varphi(p))}(mod \ p)\]

题中由于是无限层，所以答案就是 \(x\)

由于\(\varphi(\varphi(\varphi(...)))\)总有一次会变成\(1\)的，那时候\(x\%p=0\)

那么就每次递归求解\(x\%\varphi(p)\)这一块就好了啊

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define LL long long
#define max(a,b) ((a)>(b)? (a):(b))
#define min(a,b) ((a)<(b)? (a):(b))

using namespace std;

LL i,m,n,j,k,a[10001],p,t;

LL eular(LL x)
{
    LL now=x, r=sqrt(now);
    for(LL i=2;i<=r;i++) 
    {
        if(x%i!=0) continue;
        now=now/i*(i-1);
        while(x%i==0) x/=i;
    }
    if(x!=1) now=now/x*(x-1);
    return now; 
}

LL quick(LL x,LL m,LL M)
{
    LL t=1;
    for(m;m>1;m>>=1, x=x*x%M) 
        if(m&1) t=t*x%M;
    return x*t%M;
}

LL dfs(LL now)
{
    if(now==1) return 0;
    LL phi=eular(now);
    return quick(n,dfs(phi)+phi,now);
}


int main()
{
    scanf("%lld",&t);
    for(t;t;t--)
    {
        scanf("%lld%lld",&n,&p);
        printf("%lld\n",dfs(p));
    }
}

转载于:https://www.cnblogs.com/ZUTTER/p/10229858.html